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Thermodynamic Potentials Combined 1st and 2nd Law is a function of S and p H = H(S, p) dU = TdS - pdV & 'H # & 'H # dH = $ ! dS + $ ! dp U =U (S,V ) % 'S " p % 'p "S & 'U # & 'U # & 'H # & 'H # & 'T # & 'V # dU = $ ! dS + $ ! dV T = $ ! V = $ ! , $ ! = $ ! % 'S "V % 'V "S % 'S " p % 'p "S % 'p "S % 'S " p & 'U # & 'U # & 'T # & 'p # T = $ ! p = -$ ! , - $ ! = $ ! % 'S "V % 'V "S % 'V "S % 'S "V

Potential Function in terms of S and p, Enthalpy Potential Function in terms of T and p, Gibbs Free Lengendre Transform ! subtract a -d(pV) term Lengendre Transform ! add a -d(TS) term from dU to dH d ( H-TS ) = Vdp - SdT dU + d(pV) = TdS - pdV + d(pV) where G = (H-TS) is the and G = G (p, T) d(U + pV) = TdS + Vdp & 'G # & 'G # dG = $ ! dT + $ ! dp where H = (U + pV) is the Enthalpy % 'T " p % 'p "T and H = H (S, p) & 'G # & 'G # & 'S # & 'V # S = $ ! ,V = $ ! , $ ! = $ ! % 'T " p % 'p "T % 'p "T % 'T " p 1 Potential Function in terms of T and V, Four Fundamental Thermodynamic Potentials dU = TdS - pdV Lengendre Transform ! subtract a -d(TS) term from dU dH = TdS + Vdp d(U-TS) = -pdV - SdT = dA where A = A(V, T) is the dG = Vdp - SdT Helmholtz Free Energy

& 'A # & 'A # dA = $ ! dV + $ ! dT dA = -pdV - SdT % 'V "T % 'T "V & 'A # & 'A # & 'S # & 'p # S = ($ ! , p = ($ ! , $ ! = $ ! The appropriate thermodynamic potential % 'T "V % 'V "T % 'V "T % 'T "V to use is determined by the constraints imposed on the system. For example, The Maxwell relations are useful in that since is hard to control (adiabatic the relate quantities that are difficult or conditions are difficult to impose) G and A impossible to measure to quantities that are more useful. Also in the case of can be measured. p is a lot easier to control than V so G is the most useful of all potentials for solids.

2 The Thermodynamic Potentials

Four Fundamental Thermodynamic Potentials Equilibrium dU = TdS - pdV U== U( S , V ); dU 0 fixed V,S

dH = TdS + Vdp H== H( S , P ); dH 0 fixed S,P

dG = Vdp - SdT G== G( P , T ); dG 0 fixed P,T

dA = -pdV - SdT A== A( V , T ); dA 0 fixed T,V

The appropriate thermodynamic potential to use is determined by the constraints imposed on the system. For example, since entropy is hard to control (adiabatic conditions are difficult to impose) G and A are more useful. Also in the case of solids p is a lot easier to control than V so G is the most useful of all potentials for solids. The Gibbs Phase Rule

P ! C + 2 For a system of C independent components not more than C + 2 phases can co-exist in equilibrium.

Trivial Example: C = 1, P = 3 , liquid vapor

If P is less than C + 2 then C + 2 – P variables can take on arbitrary values (degrees of freedom) without disturbing equilibrium. def. Thermodynamic degrees of freedom, f

fC=+20!" P Gibbs Phase Rule of surface and interfaces

Define ! : Consider ! to be a force / unit length of surface perimeter. (fluid systems)

If a portion of the perimeter moves an infinitesimal of distance in the plane of the surface of area A, the area change dA is a product of that portion of perimeter and the length moved.

Work term - !dA; force x"distance, and appears in the combined 1st and 2nd as

dU = TdS " pdV + #µidN i +! dA i Strictly speaking , ! is defined as the change in when the area is reversibly increased at constant S, V and Ni (i.e., closed system).

For a system containing a plane surface this equation can be readily integrated :

UTSPV= " ++#µii N! A i and rearranging for ! yields.

& 1 #& # ) = $ !$U 'TS + PV ' (µi Ni ! % A "% i " where U – TS + PV is the Gibbs free energy of the system, i.e., the actual energy of the system And ! µi Ni is the Gibbs free energy of the materials comprising the i system, i.e., the energy of the system as if it were uniform ignoring any variations associated with the surface

Thus ! is an excess free energy due to the presence of the surface.

def Surface Excess Quantities

Macroscopic extensive properties of an separating bulk phases are defined as a surface excess. There is a hypothetical 2D “dividing surface” defined for which the parameters of the bulk phases change discontinuously at the dividing surface.

The excess is defined as the difference between the actual value of the def extensive quantity in the system and that which would have been present in the same if the phases were homogeneous right up to the “ Dividing Surface ” i.e., xs = xtotal # (x" + x! )

The real value of x in The values of x in the homogeneous the system ! and ! phases Concept of the Gibbs Dividing Surface

Extensive property

Density ! !

Distance perpendicular to the surface

For a 1 component system the position of the dividing surface is chosen such that the two shared areas in the figure are equal. This yields a consistent value (equal to zero ) for the surface excess. For a multicomponent system the position of the dividing surface that makes some Ni equal to zero will be unlikely to make all the other Nj ≠i = 0.

By convention, N1, the surface excess of the component present in the largest amount (i.e., the solvent) is made zero by appropriate choice of dividing surface.

Alternatively if we consider a large homogeneous crystalline body containing N atoms surrounded by plane surfaces then if U0 and S0 are the energy and entropy / per atom, the surface energy per unit area Us is defined by UNUAUNUA=+00ss" + !! : U" where U is the total energy of the system. Similarly

S = S 0 N + AS s

Consider once again the combined form of 1st and 2nd laws including the surface term.

dU = TdS " pdV + #µidN i +! dA i Substitution of the definition of G leads to

dG = "SdT +Vdp + #µidN i +! dA i If the surface is reversibly created in a closed system (Ni fixed) at constant T and P. ' (G $ % " = ! (A & #T ,P,Ni

! is always the free energy change appropriate to the constraints imposed on the system. Since for the bulk phases ! and " the surface terms vanish, the combined 1st and 2nd law take the form

!! ! ! dU= TdS" pdV+ #µii dN and i !! ! ! dU= TdS" pdV+ #µii dN i and for the total system

dU= TdS" pdV++#µii dN! dA i s !" From the definition of surface excess: XXXX= % &'#$+

sss s By Def. dU= TdS" pdV ++#µii dN ! dA =0 i Integration yields,

ss U=+ TS"µii N +! A i

Forming the Gibbs-Duhem relation :

s 0 = "S dT " # Nidµi " Ad! i so

Gibbs- Equation dsdTd! = ""#$ iiµ i

S s N s where s = !"; = i AAi Solid and liquid Surfaces In a nn pair potential model of a solid, the surface free energy can be thought of as the energy/ unit -area associated with bond breaking. :

n work/ unit area to create new surface = " # 2! A where n/A is the # of broken bonds / unit-area and the ! is the energy per bond i.e., the well depth in the pair-potential. ! Then letting A = a2 where a lattice spacing " # 2a2 pair potential If the solid is sketched such that U(r) the surface area is altered a " a + da r A " A + dA the energy ! " ! + d! !

The total energy of the surface is changed by an amount. U S = !Asurf . dU = !dA + Ad! dU d! and ==+fA! dA dA Surface Stress and Surface Energy

Unit Cube The difference in the work per unit area required for the W1=2γ constrained stretching (fix fxx dimension in the y direction while 1 Split stretching along the x-direction) is Stretch defined as the surface stress, fxx. W2 This is the excess work owing to w1 the presence of the surfaces.

fxx

w =2(γ+dγ)(1+dx) 2 1+dx

Shuttleworth cycle relating surface stress, f and surface energy, γ. Surface Stress and Surface Energy

Relation between fij and ! Consider 2 paths to get to the same final state of the deformed halves.

Path I - The cube is first stretched Path II - The cube is first separated and then separated. and then stretched.

WI = w1 + w2 WII = W1 + W2 = w1 + 2(1+dx) (γ + Δγ) = 2 γ + W2 = w1 + 2 γ + 2 Δγ + 2 γ dx where "#xx (= dx/1) has caused a change Δγ in γ.

Since WI = WII, w1 + 2 ! + 2 "! + 2 ! "#xx = 2 ! + W2 work/unit area = (W2 - w1)/2"#xx = fxx = ! + " !/"#xx Surface stress, surface free energy and chemical equilibrium of small

Recall that for finite-size liquid drops in equilibrium with the vapor. (see condensation discussion) 2! µµ=+V Equil. cond. ll0 r where Vl is the molar vol. of the liquid.

For a finite-size solid of radius r the internal pressure is a function of the size owing to the surface stress {isotropic surface stress}. 2 f µ = µ + V s 0 r s The pressure difference between the finite-size solid in equil. with the liquid is 2 f PP=+ slr

Consider the equilibrium between a solid sphere and a fluid containing the dissolved solid.

r The$total$energy$of$the$system$is$given$by

dU= TdS# pdV++$ µ dNi ! dA i =0

dU=++ TdSs() solid TdS l ( liquid ) TdS"" ( surface )### p s dV s p l dV l pdV NN ss ll ss ll +µµ11dN++ 11 dN$$ µii dN + µ ii dN +! dA ii==22 Gibbs$dividing$surface$set$for$component$1,$other$components$are$not$allowed$ to$cause$area$changes.

dU=++ TdSssolid() TdS lliquid ( ) TdS" ( surface )## p s dV s p l dV l ss ll +µµ11dN++ 11 dN! dA Consider)the)variation)dU = 0 under)the)indicated) constraints, dU = 0;

dSssolid()++ dS lliquid ( ) dS! ( surface ) =0,

dV==0 dVSl + dV; dV S =" dV l Sl S l dN111==0 dN + dN = 0; dN 11 =" dN

Making)the)substitutions

sl s dU= ""()() psls p dV+ µµ111" dN+=! dA 0

and)for)a)sphere,)!dA =) (2!/r)dVs 2! 0(= ""ppdV )+ (µµsl" ) dNdV s+ sls111r s ss since dVsso / dN11! V / N= V the

sl"#2! µµ11$ = ()ppsl$$ V o '(%&r

2 f Also since, pp! = Slr 2(f "! ) µµsl" = V 11 r o Now consider an N component solid of which components 1, ….. k are substitutional and k +1, …. N are interstitial.

Note that the addition or removal of interstitial atoms leaves AL unchanged.

Then N ! µil Nis = U s "TS s + (P + 2# / r) Vs i=1

N ! µis Nis = U s "TS s + (P + 2 f / r) Vs i=1 and N #(µis " µil )Nis =2( f "! )Vs / r i=1 For interstitial exchange : fluid !--interstitial---" solid

dAL = 0

µis = µil , i = k +1,...... N

For substitutional exchange : fluid !-- substitutional ---" solid

k "(µis ! µil )Nis =2( f !# )Vs / r, i =1....k i=1

k and defining and V / N as V the molar volume. ! Nis =N0 s 0 0 i=1

(µis ! µil ) = 2( f !" )V0 / r, i =1....k Examples of how finite – size effects alter equilibria

(1) Vapor pressure of a single – component solid

µµle=+()p RT ln/ P P e µ = µ( p ) + 2V f / r + V (P − P ) s e 0 !0#"#$l ≅0 in comparision to f term using

µs " µl = 2V0 f / r " RT ln p / pl = 2( f "! )V0 / r

RT ln p / pl = 2! V0 / r

same result as earlier (2) Solubility of a sparingly soluble single component solid :

µµil=+ i()CRTCC e ln/ e

µµis=+ i()2CVfr e 0 / using

RT lnC /Cl = 2! V0 / r

(3) Melting point of a single component solid :

µ = µ(T ) + S (T !T ) see l m l m Clausius – Clapyron 2 f µµ=+()TSTT( ! ) + V Equation smsmr o where Sl and Ss are molar . using

22!!VVT001 m Lf (TTm " ) ==,(SSls" ) = rSS( ls" ) rL f Tm

(4) Vapor pressure of a dilute interstitial component in a non-volatile matrix ( H in Fe….)

If the interstitial vaporizes as a molecule:

nx = xn or if it reacts with a vapor species, A, forming a compound AmXn

mA + nX = Am X n The of X in the vapor is related to the partial pressure

P of Xn or AmXn by RT p µxl = µ(pl )+ ln n pl and for the solid

µxs = µ(pl )+ 2Vx f / r

when Vx is the molar volume of X in the solid.

Using

RT ln p / pl = 2nVx f / r indicating that f determines the change in vapor pressure